O Level Physics Mechanics Quiz

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O-Level Physics Quiz - Mechanics

Name: ___________________________
Class: ___________________________
Date: ___________________________
Score: _______ / 40

Duration: 50 minutes
Total Marks: 40
Topic: Mechanics (Kinematics, Dynamics, Turning Effect of Forces, Pressure, Energy)

Instructions:

• Answer all 20 questions.
• Show your working clearly where calculation is required.
• Use $g = 10 \text{ m/s}^2$ unless stated otherwise.
• Write units in your final answers.

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Section A: Multiple Choice and Short Answer (Questions 1–5)

1. [1 mark] Which of the following quantities is a vector? A. Speed
B. Distance
C. Mass
D. Velocity

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2. [1 mark] A car travels 100 m in 5 s at constant speed. What is its average speed? _____ m/s

3. [1 mark] State the formula for calculating the moment of a force about a pivot.

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4. [1 mark] A box of mass 2 kg is at rest on a table. What is the magnitude of its weight? (Use $g = 10 \text{ m/s}^2$) _____ N

5. [1 mark] Pressure in a liquid increases with depth because of the weight of the ________ above.

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Section B: Structured Calculation (Questions 6–12)

6. [2 marks] A cyclist accelerates uniformly from $2 \text{ m/s}$ to $8 \text{ m/s}$ in 3 s. Calculate the acceleration.

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7. [2 marks] A force of 12 N is applied to a 3 kg block on a smooth surface. Find the acceleration of the block using $F = ma$.

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8. [2 marks] A uniform metre rule is pivoted at its centre (50 cm mark). A 4 N weight is hung at the 20 cm mark. Calculate the moment of this weight about the pivot.

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9. [2 marks] A hydraulic press has a small piston of area $0.01 \text{ m}^2$ and a large piston of area $0.5 \text{ m}^2$. A force of 20 N is applied to the small piston. Calculate the force exerted by the large piston.

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10. [2 marks] A 5 kg object is lifted vertically by 4 m. Calculate the increase in gravitational potential energy. ($g = 10 \text{ m/s}^2$)

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11. [3 marks] A car of mass 800 kg moves at a constant speed of $20 \text{ m/s}$. It then accelerates with a resultant force of 1600 N for 5 s. (a) Find the acceleration. (b) Find the final speed.

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12. [3 marks] A ball is dropped from rest from a height of 45 m. Assume $g = 10 \text{ m/s}^2$ and no air resistance. (a) How long does it take to reach the ground? (b) What is its speed just before impact?

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Section C: Data Interpretation and Extended Response (Questions 13–20)

13. [2 marks]

Image pending generation: graph for Q13.

Using the graph, calculate the acceleration of the bus between 10 s and 20 s.

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14. [2 marks] Using the same graph in Q13, calculate the total distance travelled by the bus in 30 s.

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15. [2 marks] Explain why a truck loaded with heavy cargo is more stable than the same truck empty, using the concept of centre of gravity.

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16. [3 marks] A uniform plank of length 4 m and weight 200 N is supported at both ends. A boy of weight 400 N stands 1 m from the left support. (a) Draw a labelled diagram showing forces and distances. (b) Calculate the reaction force at the left support.

Image pending generation: diagram for Q16.

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17. [3 marks] A crane lifts a 500 kg container at a constant speed of 2 m/s. ($g = 10 \text{ m/s}^2$) (a) Calculate the power output of the crane. (b) If the crane motor is 50% efficient, what is the input power required?

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18. [2 marks] State two differences between mass and weight.

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19. [3 marks] A block of mass 10 kg is pulled up a frictionless slope of height 3 m using a force parallel to the slope. The slope length is 5 m. (a) Calculate the gain in gravitational potential energy. (b) Calculate the magnitude of the pulling force.

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20. [3 marks] A student uses a lever to lift a 300 N rock. The pivot is 0.2 m from the rock and 0.8 m from the effort applied. (a) Calculate the effort needed to balance the rock. (b) State one reason why the actual effort needed is greater than your answer in (a).

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O-Level Physics Quiz - Mechanics: Answer Key

Total Marks: 40
Topic: Mechanics

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Section A

1. [1 mark] D. Velocity
Teaching note: Velocity is a vector because it has both magnitude and direction. Speed, distance, and mass are scalars (magnitude only). Common mistake: choosing speed because it is related to motion.

2. [1 mark] 20 m/s
Working: Average speed = distance / time = $100 \text{ m} / 5 \text{ s} = 20 \text{ m/s}$.
Teaching note: Remind students to use total distance over total time for average speed.

3. [1 mark] Moment = Force × perpendicular distance from pivot
Teaching note: The perpendicular distance is critical; if force is not perpendicular, use the component perpendicular to the lever arm.

4. [1 mark] 20 N
Working: Weight = $m \times g = 2 \text{ kg} \times 10 \text{ m/s}^2 = 20 \text{ N}$.
Teaching note: Weight is a force due to gravity; mass is constant but weight depends on g.

5. [1 mark] liquid (or water / fluid)
Teaching note: Pressure in liquid = $h \rho g$; greater depth means more liquid weight above exerting downward force.

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Section B

6. [2 marks]
Working: $a = (v - u)/t = (8 - 2)/3 = 6/3 = 2 \text{ m/s}^2$.
Answer: $2 \text{ m/s}^2$.
Marking: 1 mark for correct formula, 1 mark for correct substitution and answer.

7. [2 marks]
Working: $F = ma \Rightarrow a = F/m = 12 / 3 = 4 \text{ m/s}^2$.
Answer: $4 \text{ m/s}^2$.
Teaching note: Smooth surface means no friction; resultant force = applied force.

8. [2 marks]
Working: Distance from pivot = $50 - 20 = 30 \text{ cm} = 0.30 \text{ m}$.
Moment = $F \times d = 4 \times 0.30 = 1.2 \text{ Nm}$.
Answer: $1.2 \text{ Nm}$.
Marking: 1 mark for distance conversion, 1 mark for moment.

9. [2 marks]
Working: Pressure same in liquid: $F_1/A_1 = F_2/A_2$.
$F_2 = F_1 \times (A_2/A_1) = 20 \times (0.5 / 0.01) = 20 \times 50 = 1000 \text{ N}$.
Answer: 1000 N.
Teaching note: Hydraulic system multiplies force by area ratio.

10. [2 marks]
Working: $\Delta E_p = mgh = 5 \times 10 \times 4 = 200 \text{ J}$.
Answer: 200 J.
Marking: 1 mark formula, 1 mark answer.

11. [3 marks]
(a) $a = F/m = 1600 / 800 = 2 \text{ m/s}^2$ (1 mark)
(b) $v = u + at = 20 + 2 \times 5 = 30 \text{ m/s}$ (2 marks: 1 for method, 1 for answer)
Teaching note: Constant speed initially means u = 20 m/s; resultant force causes acceleration.

12. [3 marks]
(a) $s = \frac{1}{2}gt^2 \Rightarrow 45 = \frac{1}{2} \times 10 \times t^2 = 5t^2 \Rightarrow t^2 = 9 \Rightarrow t = 3 \text{ s}$ (2 marks)
(b) $v = gt = 10 \times 3 = 30 \text{ m/s}$ (1 mark)
Teaching note: Dropped from rest so u=0; free fall uses g=10.

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Section C

13. [2 marks]
Working: Acceleration = change in velocity / time = $(15 - 5) / (20 - 10) = 10 / 10 = 1 \text{ m/s}^2$.
Answer: $1 \text{ m/s}^2$.
From graph: slope of line between 10 and 20 s.

14. [2 marks]
Working: Distance = area under graph.
0–10 s: $5 \times 10 = 50 \text{ m}$
10–20 s: trapezium = $\frac{1}{2}(5+15) \times 10 = 100 \text{ m}$
20–30 s: $15 \times 10 = 150 \text{ m}$
Total = $50+100+150 = 300 \text{ m}$.
Answer: 300 m.
Marking: 1 mark for area method, 1 mark total.

15. [2 marks]
Answer points:

• Loaded truck has lower centre of gravity (cargo low). (1 mark)
• Lower CG means line of action of weight falls within base of support more easily, increasing stability. (1 mark)
  Teaching note: Stability improves when CG is low and base wide.

16. [3 marks]
(a) Diagram: show plank AB, reaction R_A up at A, R_B up at B, 200 N down at 2 m, 400 N down at 1 m from A. (1 mark)
(b) Take moments about B: $R_A \times 4 = 400 \times 3 + 200 \times 2 = 1200 + 400 = 1600 \Rightarrow R_A = 400 \text{ N}$. (2 marks)
Teaching note: Use principle of moments for equilibrium.

17. [3 marks]
(a) Constant speed → force = weight = $500 \times 10 = 5000 \text{ N}$. Power = $Fv = 5000 \times 2 = 10000 \text{ W}$. (2 marks)
(b) Efficiency = useful output / input → input = 10000 / 0.5 = 20000 W. (1 mark)
Answer: (a) 10000 W, (b) 20000 W.

18. [2 marks]

• Mass is amount of matter, weight is gravitational force. (1 mark)
• Mass is scalar and constant, weight is vector and varies with g. (1 mark)
  Other acceptable: measured in kg vs N.

19. [3 marks]
(a) $\Delta E_p = mgh = 10 \times 10 \times 3 = 300 \text{ J}$. (1 mark)
(b) Work done by force = $F \times 5 = 300 \Rightarrow F = 60 \text{ N}$. (2 marks)
Teaching note: Frictionless so all work becomes PE.

20. [3 marks]
(a) Moments: $300 \times 0.2 = E \times 0.8 \Rightarrow E = 60 / 0.8 = 75 \text{ N}$. (2 marks)
(b) Friction at pivot / weight of lever / effort not perfectly perpendicular. (1 mark)
Answer: (a) 75 N.